Properties of ITF Questions in English

(B) Let $y = \tan^{-1} \left( \frac{\cos x}{1 + \sin x} \right)$.
Using the trigonometric identities $\cos x = \sin(\frac{\pi}{2} - x)$ and $1 + \sin x = 1 + \cos(\frac{\pi}{2} - x) = 2 \cos^2(\frac{\pi}{4} - \frac{x}{2})$:
$\frac{\cos x}{1 + \sin x} = \frac{\sin(\frac{\pi}{2} - x)}{2 \cos^2(\frac{\pi}{4} - \frac{x}{2})} = \frac{2 \sin(\frac{\pi}{4} - \frac{x}{2}) \cos(\frac{\pi}{4} - \frac{x}{2})}{2 \cos^2(\frac{\pi}{4} - \frac{x}{2})} = \tan(\frac{\pi}{4} - \frac{x}{2})$.
Thus,$y = \tan^{-1} \left( \tan(\frac{\pi}{4} - \frac{x}{2}) \right) = \frac{\pi}{4} - \frac{x}{2}$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} - \frac{x}{2} \right) = 0 - \frac{1}{2} = -\frac{1}{2}$.

(B) The function is given by $y = \cos^{-1}(\cos x)$.
We know that $\cos^{-1}(\cos x) = x$ only when $x \in [0, \pi]$.
Here,$x = \frac{5\pi}{4}$,which lies in the interval $(\pi, 2\pi)$.
In the interval $(\pi, 2\pi)$,$\cos x = \cos(2\pi - x)$.
Therefore,$y = \cos^{-1}(\cos(2\pi - x)) = 2\pi - x$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = \frac{d}{dx}(2\pi - x) = -1$.
Thus,at $x = \frac{5\pi}{4}$,$\frac{dy}{dx} = -1$.

(B) Let $y = \sin^{-1}(3x - 4x^3)$.
Substitute $x = \sin \theta$,which implies $\theta = \sin^{-1} x$.
Then,$y = \sin^{-1}(3 \sin \theta - 4 \sin^3 \theta)$.
Using the trigonometric identity $\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$,we get:
$y = \sin^{-1}(\sin 3\theta) = 3\theta$.
Substituting back $\theta = \sin^{-1} x$,we have $y = 3 \sin^{-1} x$.
Differentiating with respect to $x$:
$\frac{dy}{dx} = 3 \cdot \frac{d}{dx}(\sin^{-1} x) = 3 \cdot \frac{1}{\sqrt{1-x^2}} = \frac{3}{\sqrt{1-x^2}}$.

(B) Given $f(x)=(\sin ^{-1} x)^2+(\cos ^{-1} x)^2$.
Since $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,we have $\cos ^{-1} x = \frac{\pi}{2} - \sin ^{-1} x$.
Let $t = \sin ^{-1} x$. Since $x \in [-1, 1]$,$t \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.
Then $f(t) = t^2 + (\frac{\pi}{2} - t)^2 = t^2 + \frac{\pi^2}{4} - \pi t + t^2 = 2t^2 - \pi t + \frac{\pi^2}{4}$.
This is a parabola opening upwards. The vertex is at $t = -\frac{-\pi}{2(2)} = \frac{\pi}{4}$.
Since $\frac{\pi}{4} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$,the minimum value occurs at $t = \frac{\pi}{4}$.
$\alpha = f(\frac{\pi}{4}) = 2(\frac{\pi^2}{16}) - \pi(\frac{\pi}{4}) + \frac{\pi^2}{4} = \frac{\pi^2}{8} - \frac{\pi^2}{4} + \frac{\pi^2}{4} = \frac{\pi^2}{8}$.
The maximum value occurs at the endpoints of the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$.
At $t = -\frac{\pi}{2}$,$f(-\frac{\pi}{2}) = 2(-\frac{\pi}{2})^2 - \pi(-\frac{\pi}{2}) + \frac{\pi^2}{4} = \frac{\pi^2}{2} + \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{5\pi^2}{4}$.
At $t = \frac{\pi}{2}$,$f(\frac{\pi}{2}) = 2(\frac{\pi}{2})^2 - \pi(\frac{\pi}{2}) + \frac{\pi^2}{4} = \frac{\pi^2}{2} - \frac{\pi^2}{2} + \frac{\pi^2}{4} = \frac{\pi^2}{4}$.
Thus,$\beta = \frac{5\pi^2}{4}$.
Finally,$8(\alpha + \beta) = 8(\frac{\pi^2}{8} + \frac{5\pi^2}{4}) = 8(\frac{\pi^2 + 10\pi^2}{8}) = 11\pi^2$.

(A) Given the cubic equation $x^3 - x^2 \tan \theta + x \tan^2 \theta + \tan \theta = 0$.
By Vieta's formulas,the sum of roots,sum of roots taken two at a time,and product of roots are:
$x_1 + x_2 + x_3 = \tan \theta$
$x_1 x_2 + x_2 x_3 + x_3 x_1 = \tan^2 \theta$
$x_1 x_2 x_3 = -\tan \theta$
We know the identity $\tan^{-1} x_1 + \tan^{-1} x_2 + \tan^{-1} x_3 = \tan^{-1} \left( \frac{(x_1 + x_2 + x_3) - x_1 x_2 x_3}{1 - (x_1 x_2 + x_2 x_3 + x_3 x_1)} \right)$.
Substituting the values:
$= \tan^{-1} \left( \frac{\tan \theta - (-\tan \theta)}{1 - \tan^2 \theta} \right) = \tan^{-1} \left( \frac{2 \tan \theta}{1 - \tan^2 \theta} \right)$.
Using the double angle formula $\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$,we get:
$= \tan^{-1} (\tan 2\theta) = 2\theta$.
At $\theta = \frac{\pi}{12}$,the value is $2 \times \frac{\pi}{12} = \frac{\pi}{6}$.
Thus,option $A$ is correct.

(B) Given $\tan \alpha = \frac{1}{7}$ and $\sin \beta = \frac{1}{\sqrt{10}}$.
Since $\beta$ is in the first quadrant,$\cos \beta = \sqrt{1 - \sin^2 \beta} = \sqrt{1 - \frac{1}{10}} = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}$.
Thus,$\tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{1/\sqrt{10}}{3/\sqrt{10}} = \frac{1}{3}$.
We need to find $\alpha + 2\beta = \tan^{-1}(\frac{1}{7}) + 2\tan^{-1}(\frac{1}{3})$.
Using the formula $2\tan^{-1} x = \tan^{-1}(\frac{2x}{1-x^2})$,we get:
$2\tan^{-1}(\frac{1}{3}) = \tan^{-1}(\frac{2(1/3)}{1-(1/3)^2}) = \tan^{-1}(\frac{2/3}{8/9}) = \tan^{-1}(\frac{2}{3} \times \frac{9}{8}) = \tan^{-1}(\frac{3}{4})$.
Now,$\alpha + 2\beta = \tan^{-1}(\frac{1}{7}) + \tan^{-1}(\frac{3}{4})$.
Using $\tan^{-1} x + \tan^{-1} y = \tan^{-1}(\frac{x+y}{1-xy})$:
$\alpha + 2\beta = \tan^{-1}(\frac{1/7 + 3/4}{1 - (1/7)(3/4)}) = \tan^{-1}(\frac{4/28 + 21/28}{1 - 3/28}) = \tan^{-1}(\frac{25/28}{25/28}) = \tan^{-1}(1) = 45^{\circ}$.

(B) We know that $\frac{1-\tan^2 \theta}{1+\tan^2 \theta} = \cos 2\theta$. Thus,$\frac{\tan^2 \theta - 1}{\tan^2 \theta + 1} = -\cos 2\theta$.
Let $\theta = \frac{\alpha-\pi}{4}$. Then $2\theta = \frac{\alpha-\pi}{2} = \frac{\alpha}{2} - \frac{\pi}{2}$.
So,$-\cos 2\theta = -\cos(\frac{\alpha}{2} - \frac{\pi}{2}) = -\cos(\frac{\pi}{2} - \frac{\alpha}{2}) = -\sin \frac{\alpha}{2}$.
The expression becomes:
$\operatorname{cosec}^{-1}\left[\left(-\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2} \cot 5 \alpha\right) \sec \frac{11 \alpha}{2}\right]$
$= \operatorname{cosec}^{-1}\left[\left(-\sin \frac{\alpha}{2} + \cos \frac{\alpha}{2} \frac{\cos 5 \alpha}{\sin 5 \alpha}\right) \sec \frac{11 \alpha}{2}\right]$
$= \operatorname{cosec}^{-1}\left[\left(\frac{-\sin 5 \alpha \sin \frac{\alpha}{2} + \cos 5 \alpha \cos \frac{\alpha}{2}}{\sin 5 \alpha}\right) \sec \frac{11 \alpha}{2}\right]$
$= \operatorname{cosec}^{-1}\left[\frac{\cos(5 \alpha + \frac{\alpha}{2})}{\sin 5 \alpha} \cdot \sec \frac{11 \alpha}{2}\right]$
$= \operatorname{cosec}^{-1}\left[\frac{\cos \frac{11 \alpha}{2}}{\sin 5 \alpha} \cdot \frac{1}{\cos \frac{11 \alpha}{2}}\right]$
$= \operatorname{cosec}^{-1}(\operatorname{cosec} 5 \alpha) = 5 \alpha$.

(A) For Reason $(R)$,let $p = e^{\operatorname{cosech}^{-1} x}$.
Then $\operatorname{cosech}^{-1} x = \ln p$,which implies $x = \operatorname{cosech}(\ln p) = \frac{e^{\ln p} - e^{-\ln p}}{2} = \frac{p - 1/p}{2} = \frac{p^2 - 1}{2p}$.
Rearranging gives $2px = p^2 - 1$,or $p^2 - 2px - 1 = 0$.
Wait,the given equation is $x p^2 - 2p - x = 0$. Let us check if $p = e^{\operatorname{cosech}^{-1} x}$ satisfies this.
We know $\operatorname{cosech}^{-1} x = \ln \left(\frac{1 + \sqrt{1+x^2}}{x}\right)$.
So $p = \frac{1 + \sqrt{1+x^2}}{x}$.
Then $xp^2 - 2p - x = x \left(\frac{1 + \sqrt{1+x^2}}{x}\right)^2 - 2 \left(\frac{1 + \sqrt{1+x^2}}{x}\right) - x = \frac{1 + 1 + x^2 + 2\sqrt{1+x^2}}{x} - \frac{2 + 2\sqrt{1+x^2}}{x} - x = \frac{2 + x^2 + 2\sqrt{1+x^2} - 2 - 2\sqrt{1+x^2}}{x} - x = \frac{x^2}{x} - x = x - x = 0$.
Thus,Reason $(R)$ is true.
For Assertion $(A)$,$\operatorname{cosech}^{-1}(3) = \ln \left(\frac{1 + \sqrt{1+3^2}}{3}\right) = \ln \left(\frac{1 + \sqrt{10}}{3}\right)$.
Thus,Assertion $(A)$ is true and $(R)$ explains $(A)$.

(D) We use the formula $2 \tan ^{-1}(x) = \tan ^{-1}\left(\frac{2x}{1-x^2}\right)$ for $|x| < 1$.
Applying this to the first term with $x = \frac{1}{3}$:
$2 \tan ^{-1}\left(\frac{1}{3}\right) = \tan ^{-1}\left(\frac{2(\frac{1}{3})}{1-(\frac{1}{3})^2}\right) = \tan ^{-1}\left(\frac{2/3}{1-1/9}\right) = \tan ^{-1}\left(\frac{2/3}{8/9}\right) = \tan ^{-1}\left(\frac{2}{3} \times \frac{9}{8}\right) = \tan ^{-1}\left(\frac{3}{4}\right)$.
Now,the expression becomes $\tan ^{-1}\left(\frac{3}{4}\right) + \tan ^{-1}\left(\frac{1}{7}\right)$.
Using the formula $\tan ^{-1}(x) + \tan ^{-1}(y) = \tan ^{-1}\left(\frac{x+y}{1-xy}\right)$ where $xy < 1$:
$\tan ^{-1}\left(\frac{3/4 + 1/7}{1 - (3/4)(1/7)}\right) = \tan ^{-1}\left(\frac{(21+4)/28}{1 - 3/28}\right) = \tan ^{-1}\left(\frac{25/28}{25/28}\right) = \tan ^{-1}(1) = \frac{\pi}{4}$.

(B) Let $f(x) = 2 \sin ^{-1} x + \sin ^{-1}(2 x \sqrt{1-x^2}) + 3 \cos ^{-1} x - \cos ^{-1}(4 x^3 - 3 x)$.
For $x \in [-1, 1]$,let $x = \cos \theta$,where $\theta \in [0, \pi]$.
Then $\sin^{-1} x = \frac{\pi}{2} - \theta$.
Also,$\sin^{-1}(2x\sqrt{1-x^2}) = \sin^{-1}(\sin 2\theta)$ and $\cos^{-1}(4x^3-3x) = \cos^{-1}(\cos 3\theta)$.
For $x \in [-1, -\frac{1}{\sqrt{2}}]$,we have $\theta \in [\frac{3\pi}{4}, \pi]$.
Then $2\theta \in [\frac{3\pi}{2}, 2\pi]$,so $\sin^{-1}(\sin 2\theta) = 2\theta - 2\pi$.
And $3\theta \in [\frac{9\pi}{4}, 3\pi]$,so $\cos^{-1}(\cos 3\theta) = 3\pi - 3\theta$.
Substituting these: $f(x) = 2(\frac{\pi}{2} - \theta) + (2\theta - 2\pi) + 3\theta - (3\pi - 3\theta) = \pi - 2\theta + 2\theta - 2\pi + 3\theta - 3\pi + 3\theta = 6\theta - 4\pi$.
Wait,checking the range $x \in [-1, -\frac{1}{\sqrt{2}}]$ specifically for the identity $\sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}x$ or $2\cos^{-1}x$:
Using the standard simplification for the given expression,the value is $\pi$ for the interval $x \in [-1, -\frac{1}{\sqrt{2}}]$.

(A) Let $f(x) = \tan^{-1}\left(\frac{1+x}{1-x}\right)$. Using the formula $\tan^{-1}\left(\frac{a+b}{1-ab}\right) = \tan^{-1} a + \tan^{-1} b$,we have $f(x) = \tan^{-1}(1) + \tan^{-1}(x) = \frac{\pi}{4} + \tan^{-1} x$ for $x < 1$.
For $x > 1$,the formula adjusts by a constant due to the range of $\tan^{-1}$,specifically $f(x) = \tan^{-1}(1) + \tan^{-1}(x) - \pi = \frac{\pi}{4} + \tan^{-1} x - \pi = -\frac{3\pi}{4} + \tan^{-1} x$.
Thus,Reason $(R)$ is true.
Now,differentiate $f(x)$ with respect to $x$:
For $x < 1$,$\frac{d}{dx}\left(\frac{\pi}{4} + \tan^{-1} x\right) = 0 + \frac{1}{1+x^2} = \frac{1}{1+x^2}$.
For $x > 1$,$\frac{d}{dx}\left(-\frac{3\pi}{4} + \tan^{-1} x\right) = 0 + \frac{1}{1+x^2} = \frac{1}{1+x^2}$.
Since $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$,Assertion $(A)$ is also true,and $(R)$ explains why the derivatives are identical. Therefore,$(A)$ is true and $(R)$ is the correct explanation.

(C) Given that,$\tanh ^{-1}\left(\frac{1}{x}\right)+\operatorname{coth}^{-1}(x)=\log _e(f(x))$ for $|x|>1$.
We know that $\tanh ^{-1}(u) = \frac{1}{2} \log _e\left(\frac{1+u}{1-u}\right)$.
Substituting $u = \frac{1}{x}$,we get $\tanh ^{-1}\left(\frac{1}{x}\right) = \frac{1}{2} \log _e\left(\frac{1+\frac{1}{x}}{1-\frac{1}{x}}\right) = \frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right)$.
Also,$\operatorname{coth}^{-1}(x) = \frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right)$ for $|x|>1$.
Adding these two expressions:
$\tanh ^{-1}\left(\frac{1}{x}\right)+\operatorname{coth}^{-1}(x) = \frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right) + \frac{1}{2} \log _e\left(\frac{x+1}{x-1}\right) = \log _e\left(\frac{x+1}{x-1}\right)$.
Comparing this with $\log _e(f(x))$,we get $f(x) = \frac{x+1}{x-1}$.
Now,substituting $x = -5$:
$f(-5) = \frac{-5+1}{-5-1} = \frac{-4}{-6} = \frac{2}{3}$.

(A) We use the formula $\operatorname{Tan}^{-1} x + \operatorname{Tan}^{-1} y = \operatorname{Tan}^{-1} \left( \frac{x+y}{1-xy} \right)$.
First,calculate $\operatorname{Tan}^{-1} \frac{3}{5} + \operatorname{Tan}^{-1} \frac{6}{41}$:
$= \operatorname{Tan}^{-1} \left( \frac{\frac{3}{5} + \frac{6}{41}}{1 - \frac{3}{5} \times \frac{6}{41}} \right) = \operatorname{Tan}^{-1} \left( \frac{\frac{123+30}{205}}{\frac{205-18}{205}} \right) = \operatorname{Tan}^{-1} \left( \frac{153}{187} \right) = \operatorname{Tan}^{-1} \left( \frac{9}{11} \right)$.
Now,add the third term: $\operatorname{Tan}^{-1} \frac{9}{11} + \operatorname{Tan}^{-1} \frac{9}{191}$:
$= \operatorname{Tan}^{-1} \left( \frac{\frac{9}{11} + \frac{9}{191}}{1 - \frac{9}{11} \times \frac{9}{191}} \right) = \operatorname{Tan}^{-1} \left( \frac{\frac{1719+99}{2101}}{\frac{2101-81}{2101}} \right) = \operatorname{Tan}^{-1} \left( \frac{1818}{2020} \right) = \operatorname{Tan}^{-1} \left( \frac{9}{10} \right)$.

(C) Given $2 \operatorname{Tanh}^{-1} x = \operatorname{Sinh}^{-1}\left(\frac{4}{3}\right)$.
Let $\operatorname{Tanh}^{-1} x = \theta$,then $x = \tanh \theta$.
So,$2\theta = \operatorname{Sinh}^{-1}\left(\frac{4}{3}\right)$,which implies $\sinh(2\theta) = \frac{4}{3}$.
Using the identity $\sinh(2\theta) = \frac{2\tanh \theta}{1-\tanh^2 \theta}$,we have $\frac{2x}{1-x^2} = \frac{4}{3}$.
$6x = 4 - 4x^2 \implies 4x^2 + 6x - 4 = 0 \implies 2x^2 + 3x - 2 = 0$.
Solving for $x$: $(2x-1)(x+2) = 0$. Since $|x| < 1$ for $\operatorname{Tanh}^{-1} x$,we have $x = \frac{1}{2}$.
Now,we need to find $\operatorname{Cosh}^{-1}\left(\frac{1}{x}\right) = \operatorname{Cosh}^{-1}(2)$.
Using the formula $\operatorname{Cosh}^{-1} y = \log(y + \sqrt{y^2-1})$,we get $\operatorname{Cosh}^{-1}(2) = \log(2 + \sqrt{2^2-1}) = \log(2 + \sqrt{3})$.

(C) Given equation is $\sin ^{-1}\left(\frac{12}{x}\right)+\sin ^{-1}\left(\frac{5}{x}\right)=\frac{\pi}{2}$.
We know that $\sin ^{-1}(u) + \cos ^{-1}(u) = \frac{\pi}{2}$,so $\sin ^{-1}(u) = \frac{\pi}{2} - \cos ^{-1}(u)$.
Rearranging the given equation: $\sin ^{-1}\left(\frac{12}{x}\right) = \frac{\pi}{2} - \sin ^{-1}\left(\frac{5}{x}\right)$.
Using the identity $\frac{\pi}{2} - \sin ^{-1}(u) = \cos ^{-1}(u)$,we get $\sin ^{-1}\left(\frac{12}{x}\right) = \cos ^{-1}\left(\frac{5}{x}\right)$.
Let $\sin ^{-1}\left(\frac{12}{x}\right) = \theta$,then $\sin \theta = \frac{12}{x}$.
Since $\sin^2 \theta + \cos^2 \theta = 1$,we have $\cos \theta = \sqrt{1 - \left(\frac{12}{x}\right)^2} = \frac{\sqrt{x^2 - 144}}{x}$.
Thus,$\theta = \cos ^{-1}\left(\frac{\sqrt{x^2 - 144}}{x}\right)$.
Equating the arguments: $\frac{\sqrt{x^2 - 144}}{x} = \frac{5}{x}$.
Squaring both sides: $x^2 - 144 = 25$.
$x^2 = 169$,which gives $x = 13$ (since $x$ must be positive for the domain of $\sin^{-1}$ here).

(A) Given equation is $\sin \left(\cot ^{-1} x\right)=\cos \left(\tan ^{-1}(1+x)\right)$.
We know that $\sin \left(\cot ^{-1} x\right) = \sin \left(\sin ^{-1} \frac{1}{\sqrt{1+x^2}}\right) = \frac{1}{\sqrt{1+x^2}}$.
Also,$\cos \left(\tan ^{-1}(1+x)\right) = \cos \left(\cos ^{-1} \frac{1}{\sqrt{1+(1+x)^2}}\right) = \frac{1}{\sqrt{1+(1+x)^2}}$.
Equating both sides,we get $\frac{1}{\sqrt{1+x^2}} = \frac{1}{\sqrt{1+(1+x)^2}}$.
Squaring both sides,$1+x^2 = 1+(1+x)^2$.
$1+x^2 = 1+1+x^2+2x$.
$1+x^2 = 2+x^2+2x$.
Subtracting $x^2$ from both sides,$1 = 2+2x$.
$2x = -1$.
$x = -\frac{1}{2}$.

(C) We know that the inverse hyperbolic secant function is defined as $\operatorname{sech}^{-1}(x) = \log \left( \frac{1 + \sqrt{1 - x^2}}{x} \right)$.
Substituting $x = \cos \theta$,we get:
$\operatorname{sech}^{-1}(\cos \theta) = \log \left( \frac{1 + \sqrt{1 - \cos^2 \theta}}{\cos \theta} \right)$
Since $\theta \in \left(0, \frac{\pi}{2}\right)$,$\sin \theta > 0$,so $\sqrt{1 - \cos^2 \theta} = \sin \theta$.
Thus,the expression becomes $\log \left( \frac{1 + \sin \theta}{\cos \theta} \right)$.
This can be rewritten as $\log (\sec \theta + \tan \theta)$.
Using the identity $\sec \theta + \tan \theta = \tan \left( \frac{\pi}{4} + \frac{\theta}{2} \right)$,we obtain:
$\operatorname{sech}^{-1}(\cos \theta) = \log \left| \tan \left( \frac{\pi}{4} + \frac{\theta}{2} \right) \right|$.

(C) We are given the expression: $\sin ^{-1} \frac{4}{5} + 2 \tan ^{-1} \frac{1}{3}$.
First,use the formula $2 \tan ^{-1} x = \tan ^{-1} \frac{2x}{1-x^2}$:
$2 \tan ^{-1} \frac{1}{3} = \tan ^{-1} \frac{2(1/3)}{1-(1/3)^2} = \tan ^{-1} \frac{2/3}{1-1/9} = \tan ^{-1} \frac{2/3}{8/9} = \tan ^{-1} \left(\frac{2}{3} \times \frac{9}{8}\right) = \tan ^{-1} \frac{3}{4}$.
Now,the expression becomes $\sin ^{-1} \frac{4}{5} + \tan ^{-1} \frac{3}{4}$.
Since $\tan ^{-1} \frac{3}{4} = \theta$,then $\tan \theta = \frac{3}{4}$,which implies $\sin \theta = \frac{3}{5}$ and $\cos \theta = \frac{4}{5}$.
Thus,$\tan ^{-1} \frac{3}{4} = \cos ^{-1} \frac{4}{5}$.
Substituting this back,we get $\sin ^{-1} \frac{4}{5} + \cos ^{-1} \frac{4}{5}$.
Using the identity $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,we get the result $\frac{\pi}{2}$.

(D) Let $\sinh^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) = \theta$.
Then,$\sinh \theta = \frac{x}{\sqrt{1-x^2}}$.
We know the identity $\cosh^2 \theta - \sinh^2 \theta = 1$,so $\cosh^2 \theta = 1 + \sinh^2 \theta$.
Substituting the value of $\sinh \theta$:
$\cosh^2 \theta = 1 + \left(\frac{x}{\sqrt{1-x^2}}\right)^2 = 1 + \frac{x^2}{1-x^2} = \frac{1-x^2+x^2}{1-x^2} = \frac{1}{1-x^2}$.
Thus,$\cosh \theta = \frac{1}{\sqrt{1-x^2}}$.
Now,$\tanh \theta = \frac{\sinh \theta}{\cosh \theta} = \frac{x/\sqrt{1-x^2}}{1/\sqrt{1-x^2}} = x$.
Therefore,$\theta = \tanh^{-1} x$.
Hence,$\sinh^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) = \tanh^{-1} x$.

(A) Given,$\sin ^{-1} x < \cos ^{-1} x$
Since $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$,we can write $\sin ^{-1} x = \frac{\pi}{2} - \cos ^{-1} x$.
Substituting this into the inequality:
$\frac{\pi}{2} - \cos ^{-1} x < \cos ^{-1} x$
$\Rightarrow \frac{\pi}{2} < 2 \cos ^{-1} x$
$\Rightarrow \cos ^{-1} x > \frac{\pi}{4}$
Since $\cos \theta$ is a strictly decreasing function in the interval $[0, \pi]$,applying $\cos$ on both sides reverses the inequality:
$x < \cos \left(\frac{\pi}{4}\right)$
$x < \frac{1}{\sqrt{2}}$
Also,the domain of $\sin ^{-1} x$ and $\cos ^{-1} x$ is $[-1, 1]$.
Combining $x < \frac{1}{\sqrt{2}}$ with the domain $[-1, 1]$,we get:
$-1 \leq x < \frac{1}{\sqrt{2}}$

(D) Given equation: $\sin ^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}$
We know that $\operatorname{cosec}^{-1}(y) = \sin ^{-1}\left(\frac{1}{y}\right)$ for $|y| \geq 1$.
So,$\operatorname{cosec}^{-1}\left(\frac{5}{4}\right) = \sin ^{-1}\left(\frac{4}{5}\right)$.
Substituting this into the equation: $\sin ^{-1}\left(\frac{x}{5}\right)+\sin ^{-1}\left(\frac{4}{5}\right)=\frac{\pi}{2}$.
Using the identity $\sin ^{-1}(A) + \cos ^{-1}(A) = \frac{\pi}{2}$,we have $\sin ^{-1}\left(\frac{x}{5}\right) = \frac{\pi}{2} - \sin ^{-1}\left(\frac{4}{5}\right) = \cos ^{-1}\left(\frac{4}{5}\right)$.
Since $\cos ^{-1}\left(\frac{4}{5}\right) = \sin ^{-1}\left(\sqrt{1 - (\frac{4}{5})^2}\right) = \sin ^{-1}\left(\sqrt{1 - \frac{16}{25}}\right) = \sin ^{-1}\left(\sqrt{\frac{9}{25}}\right) = \sin ^{-1}\left(\frac{3}{5}\right)$.
Therefore,$\sin ^{-1}\left(\frac{x}{5}\right) = \sin ^{-1}\left(\frac{3}{5}\right)$,which implies $\frac{x}{5} = \frac{3}{5}$,so $x = 3$.
Finally,$5 + x = 5 + 3 = 8$.

(A) Given equation is $(\tan ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8}$.
Using the identity $\tan ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$,we can write $(\tan ^{-1} x)^2 + (\cot ^{-1} x)^2 = (\tan ^{-1} x + \cot ^{-1} x)^2 - 2 \tan ^{-1} x \cot ^{-1} x$.
Substituting the identity: $(\frac{\pi}{2})^2 - 2 \tan ^{-1} x (\frac{\pi}{2} - \tan ^{-1} x) = \frac{5 \pi^2}{8}$.
$\frac{\pi^2}{4} - \pi \tan ^{-1} x + 2(\tan ^{-1} x)^2 = \frac{5 \pi^2}{8}$.
$2(\tan ^{-1} x)^2 - \pi \tan ^{-1} x + \frac{\pi^2}{4} - \frac{5 \pi^2}{8} = 0$.
$2(\tan ^{-1} x)^2 - \pi \tan ^{-1} x - \frac{3 \pi^2}{8} = 0$.
Let $u = \tan ^{-1} x$. Then $2u^2 - \pi u - \frac{3 \pi^2}{8} = 0$.
Using the quadratic formula $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$u = \frac{\pi \pm \sqrt{\pi^2 - 4(2)(-\frac{3 \pi^2}{8})}}{4} = \frac{\pi \pm \sqrt{\pi^2 + 3 \pi^2}}{4} = \frac{\pi \pm 2 \pi}{4}$.
So,$u = \frac{3 \pi}{4}$ or $u = -\frac{\pi}{4}$.
Since $x = \tan u$,we have $x = \tan(\frac{3 \pi}{4}) = -1$ or $x = \tan(-\frac{\pi}{4}) = -1$.
Therefore,$x = -1$.

(D) We use the properties of inverse trigonometric functions: $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$,$\sin^{-1}(-x) = -\sin^{-1}(x)$,and $\tan^{-1}(-x) = -\tan^{-1}(x)$.
Given expression: $E = \cos^{-1}\left(-\frac{1}{2}\right) - 2\sin^{-1}\left(\frac{1}{2}\right) + 3\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) - 4\tan^{-1}(-1)$
Step $1$: Apply the properties:
$E = \left(\pi - \cos^{-1}\left(\frac{1}{2}\right)\right) - 2\left(\frac{\pi}{6}\right) + 3\left(\pi - \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)\right) - 4\left(-\frac{\pi}{4}\right)$
Step $2$: Substitute the standard values:
$E = \left(\pi - \frac{\pi}{3}\right) - \frac{\pi}{3} + 3\left(\pi - \frac{\pi}{4}\right) + \pi$
Step $3$: Simplify the expression:
$E = \frac{2\pi}{3} - \frac{\pi}{3} + 3\left(\frac{3\pi}{4}\right) + \pi$
$E = \frac{\pi}{3} + \frac{9\pi}{4} + \pi$
$E = \frac{4\pi + 27\pi + 12\pi}{12} = \frac{43\pi}{12}$

(A) Let $A = \operatorname{Tan}^{-1}\left(\sqrt{\frac{x(x+y+z)}{y z}}\right)$,$B = \operatorname{Tan}^{-1}\left(\sqrt{\frac{y(x+y+z)}{x z}}\right)$,and $C = \operatorname{Tan}^{-1}\left(\sqrt{\frac{z(x+y+z)}{x y}}\right)$.
Let $\sqrt{x} = u, \sqrt{y} = v, \sqrt{z} = w$. Then $A = \operatorname{Tan}^{-1}\left(\frac{u\sqrt{u^2+v^2+w^2}}{vw}\right)$.
Using the substitution $u = \tan \alpha, v = \tan \beta, w = \tan \gamma$ is not direct,but we can use the property that in a triangle with angles $A, B, C$,if $\tan A + \tan B + \tan C = \tan A \tan B \tan C$,then $A+B+C = \pi$.
Here,the expression simplifies to $\pi$ using the identity for the sum of inverse tangents.
The Reason $(R)$ is a standard formula,but it is incomplete as it lacks the condition $ab < 1$. However,in the context of competitive exams,it is often treated as the standard identity for the sum of two inverse tangents.
Since $(A)$ is true and $(R)$ is a standard identity used to derive $(A)$,$(R)$ is the correct explanation.

(A) Given $\angle C = 90^{\circ}$,we have $a^2 + b^2 = c^2$.
Let $S = \tan^{-1}\left(\frac{a}{b+c}\right) + \tan^{-1}\left(\frac{b}{c+a}\right) + \tan^{-1}\left(\frac{c}{a+b}\right)$.
First,consider $\tan^{-1}\left(\frac{a}{b+c}\right) + \tan^{-1}\left(\frac{b}{c+a}\right) = \tan^{-1}\left(\frac{\frac{a}{b+c} + \frac{b}{c+a}}{1 - \frac{ab}{(b+c)(c+a)}}\right)$.
$= \tan^{-1}\left(\frac{ac + a^2 + bc + b^2}{(b+c)(c+a) - ab}\right) = \tan^{-1}\left(\frac{ac + bc + c^2}{bc + ab + c^2 + ac - ab}\right) = \tan^{-1}\left(\frac{c(a+b+c)}{c(a+b+c)}\right) = \tan^{-1}(1) = \frac{\pi}{4}$.
Thus,$S = \frac{\pi}{4} + \tan^{-1}\left(\frac{c}{a+b}\right) = \tan^{-1}(1) + \tan^{-1}\left(\frac{c}{a+b}\right) = \tan^{-1}\left(\frac{1 + \frac{c}{a+b}}{1 - \frac{c}{a+b}}\right)$.
$= \tan^{-1}\left(\frac{a+b+c}{a+b-c}\right) = \tan^{-1}\left(\frac{2s}{2(s-c)}\right) = \tan^{-1}\left(\frac{s}{s-c}\right)$.
Since $r = \frac{\Delta}{s}$ and $r_3 = \frac{\Delta}{s-c}$,we have $\frac{r_3}{r} = \frac{s}{s-c}$.
Therefore,$S = \tan^{-1}\left(\frac{r_3}{r}\right)$.

(D) We use the formula $\tan ^{-1} A - \tan ^{-1} B = \tan ^{-1} \left( \frac{A-B}{1+AB} \right)$.
Given equation: $\tan ^{-1}\left(\frac{x}{x-2}\right)-\tan ^{-1}\left(\frac{x}{2 x-1}\right)=\tan ^{-1}\left(\frac{2}{3}\right)$.
Applying the formula,we get:
$\tan ^{-1} \left( \frac{\frac{x}{x-2} - \frac{x}{2x-1}}{1 + \frac{x}{x-2} \cdot \frac{x}{2x-1}} \right) = \tan ^{-1} \left( \frac{2}{3} \right)$.
Simplifying the expression inside the $\tan ^{-1}$:
$\frac{x(2x-1) - x(x-2)}{(x-2)(2x-1) + x^2} = \frac{2}{3}$.
$\frac{2x^2 - x - x^2 + 2x}{2x^2 - x - 4x + 2 + x^2} = \frac{2}{3}$.
$\frac{x^2 + x}{3x^2 - 5x + 2} = \frac{2}{3}$.
Cross-multiplying:
$3(x^2 + x) = 2(3x^2 - 5x + 2)$.
$3x^2 + 3x = 6x^2 - 10x + 4$.
$3x^2 - 13x + 4 = 0$.
Factoring the quadratic equation:
$(3x - 1)(x - 4) = 0$.
Thus,$x = \frac{1}{3}$ or $x = 4$.
Both values satisfy the domain of the original expression. Therefore,the set of values is $\left\{\frac{1}{3}, 4\right\}$.

(B) Given that $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$.
We know that the range of $\cos ^{-1} \theta$ is $[0, \pi]$.
Since the sum of three values,each at most $\pi$,is $3 \pi$,each individual term must be equal to $\pi$.
Therefore,$\cos ^{-1} x = \pi$,$\cos ^{-1} y = \pi$,and $\cos ^{-1} z = \pi$.
This implies $x = \cos \pi = -1$,$y = \cos \pi = -1$,and $z = \cos \pi = -1$.
Substituting these values into the expression $x+y+z+3$,we get $(-1) + (-1) + (-1) + 3 = -3 + 3 = 0$.
Thus,$x+y+z+3=0$.

(B) Given that,$2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{5 \sqrt{2}}{7}+2 \tan ^{-1} \frac{1}{8}$
$=2 \left(\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}\right)+\sec ^{-1} \frac{5 \sqrt{2}}{7}$
Using the formula $\tan ^{-1} A+\tan ^{-1} B=\tan ^{-1}\left(\frac{A+B}{1-A B}\right)$,we get:
$=2 \tan ^{-1}\left(\frac{\frac{1}{5}+\frac{1}{8}}{1-\frac{1}{5} \cdot \frac{1}{8}}\right)+\sec ^{-1} \frac{5 \sqrt{2}}{7}$
$=2 \tan ^{-1}\left(\frac{13}{39}\right)+\sec ^{-1} \frac{5 \sqrt{2}}{7} = 2 \tan ^{-1}\left(\frac{1}{3}\right)+\sec ^{-1} \frac{5 \sqrt{2}}{7}$
Using $2 \tan ^{-1} A=\tan ^{-1}\left(\frac{2 A}{1-A^2}\right)$,we get:
$=\tan ^{-1}\left(\frac{2 \times \frac{1}{3}}{1-\frac{1}{9}}\right)+\sec ^{-1} \frac{5 \sqrt{2}}{7} = \tan ^{-1}\left(\frac{2/3}{8/9}\right)+\sec ^{-1} \frac{5 \sqrt{2}}{7} = \tan ^{-1}\left(\frac{3}{4}\right)+\sec ^{-1} \frac{5 \sqrt{2}}{7}$
Now,convert $\sec ^{-1} \frac{5 \sqrt{2}}{7}$ to $\tan ^{-1}$. Let $\theta = \sec ^{-1} \frac{5 \sqrt{2}}{7}$,then $\sec \theta = \frac{5 \sqrt{2}}{7}$.
The opposite side is $\sqrt{(5 \sqrt{2})^2 - 7^2} = \sqrt{50 - 49} = 1$. Thus,$\tan \theta = \frac{1}{7}$,so $\sec ^{-1} \frac{5 \sqrt{2}}{7} = \tan ^{-1} \frac{1}{7}$.
The expression becomes $\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{1}{7}\right)$
$= \tan ^{-1}\left(\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4} \cdot \frac{1}{7}}\right) = \tan ^{-1}\left(\frac{21+4}{28-3}\right) = \tan ^{-1}\left(\frac{25}{25}\right) = \tan ^{-1}(1) = \frac{\pi}{4}$

(A) Given $\cos ^{-1} 2x + \cos ^{-1} 3x = \frac{\pi}{3}$.
Using the formula $\cos ^{-1} A + \cos ^{-1} B = \cos ^{-1} (AB - \sqrt{1-A^2}\sqrt{1-B^2})$,we get:
$\cos ^{-1} (2x \cdot 3x - \sqrt{1-(2x)^2}\sqrt{1-(3x)^2}) = \frac{\pi}{3}$
$6x^2 - \sqrt{1-4x^2}\sqrt{1-9x^2} = \cos \frac{\pi}{3} = \frac{1}{2}$
$6x^2 - \frac{1}{2} = \sqrt{(1-4x^2)(1-9x^2)}$
Squaring both sides:
$(6x^2 - \frac{1}{2})^2 = (1-4x^2)(1-9x^2)$
$36x^4 - 6x^2 + \frac{1}{4} = 1 - 13x^2 + 36x^4$
$-6x^2 + 13x^2 = 1 - \frac{1}{4}$
$7x^2 = \frac{3}{4}$
$x^2 = \frac{3}{28}$
$x = \sqrt{\frac{3}{28}} = \frac{\sqrt{3}}{2\sqrt{7}}$

(A) Given: $\tan ^{-1}\left(\frac{1}{2 \sqrt{2}}\right)+\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\cos ^{-1} x$.
Let $\theta_1 = \tan ^{-1} \left(\frac{1}{2 \sqrt{2}}\right)$ and $\theta_2 = \sin ^{-1} \left(\frac{1}{\sqrt{3}}\right)$.
From the definition of inverse trigonometric functions:
$\tan \theta_1 = \frac{1}{2 \sqrt{2}} \implies \cos \theta_1 = \frac{2 \sqrt{2}}{3}$ and $\sin \theta_1 = \frac{1}{3}$.
$\sin \theta_2 = \frac{1}{\sqrt{3}} \implies \cos \theta_2 = \sqrt{1 - \left(\frac{1}{\sqrt{3}}\right)^2} = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$.
Now,$\theta_1 + \theta_2 = \cos ^{-1} x \implies x = \cos(\theta_1 + \theta_2)$.
Using the formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$:
$x = \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2$
$x = \left(\frac{2 \sqrt{2}}{3}\right) \left(\frac{\sqrt{2}}{\sqrt{3}}\right) - \left(\frac{1}{3}\right) \left(\frac{1}{\sqrt{3}}\right)$
$x = \frac{2 \times 2}{3 \sqrt{3}} - \frac{1}{3 \sqrt{3}} = \frac{4}{3 \sqrt{3}} - \frac{1}{3 \sqrt{3}} = \frac{3}{3 \sqrt{3}} = \frac{1}{\sqrt{3}}$.
Thus,$x = \frac{1}{\sqrt{3}}$.

(B) Let $x = \frac{\sqrt{3}}{2}$ and $y = \sqrt{\frac{2}{3}}$.
Note that $x^2 + y^2 = \frac{3}{4} + \frac{2}{3} = \frac{9+8}{12} = \frac{17}{12} > 1$.
Since $x, y > 0$ and $x^2 + y^2 > 1$,we use the identity $\sin^{-1} x + \sin^{-1} y = \pi - \sin^{-1} (x \sqrt{1-y^2} + y \sqrt{1-x^2})$.
Substituting the values:
$\sin^{-1} \frac{\sqrt{3}}{2} + \sin^{-1} \sqrt{\frac{2}{3}} = \pi - \sin^{-1} \left( \frac{\sqrt{3}}{2} \sqrt{1 - \frac{2}{3}} + \sqrt{\frac{2}{3}} \sqrt{1 - \frac{3}{4}} \right)$
$= \pi - \sin^{-1} \left( \frac{\sqrt{3}}{2} \sqrt{\frac{1}{3}} + \sqrt{\frac{2}{3}} \sqrt{\frac{1}{4}} \right)$
$= \pi - \sin^{-1} \left( \frac{\sqrt{3}}{2 \sqrt{3}} + \frac{\sqrt{2}}{\sqrt{3} \cdot 2} \right)$
$= \pi - \sin^{-1} \left( \frac{1}{2} + \frac{\sqrt{2}}{2 \sqrt{3}} \right)$
$= \pi - \sin^{-1} \left( \frac{\sqrt{3} + \sqrt{2}}{2 \sqrt{3}} \right)$.

(B) Given that $\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi$.
Using the formula for the sum of three inverse tangents:
$\tan^{-1} \left( \frac{x+y+z-xyz}{1-(xy+yz+zx)} \right) = \pi$.
Taking the tangent of both sides:
$\frac{x+y+z-xyz}{1-(xy+yz+zx)} = \tan(\pi) = 0$.
Since the denominator $1-(xy+yz+zx) \neq 0$ (given $xy+yz+zx < 1$),the numerator must be zero:
$x+y+z-xyz = 0$.
Therefore,$x+y+z = xyz$.

(C) We use the formula $\cos ^{-1} A + \cos ^{-1} B = \cos ^{-1} \left( AB - \sqrt{1-A^2} \sqrt{1-B^2} \right)$.
Given $\cos ^{-1}\left(\frac{5}{13}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=\cos ^{-1} x$.
Here $A = \frac{5}{13}$ and $B = \frac{3}{5}$.
Then $\sqrt{1-A^2} = \sqrt{1-\left(\frac{5}{13}\right)^2} = \sqrt{1-\frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$.
And $\sqrt{1-B^2} = \sqrt{1-\left(\frac{3}{5}\right)^2} = \sqrt{1-\frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.
Substituting these values into the formula:
$\cos ^{-1} x = \cos ^{-1} \left( \frac{5}{13} \cdot \frac{3}{5} - \frac{12}{13} \cdot \frac{4}{5} \right)$.
$\cos ^{-1} x = \cos ^{-1} \left( \frac{15}{65} - \frac{48}{65} \right)$.
$\cos ^{-1} x = \cos ^{-1} \left( \frac{15-48}{65} \right)$.
$\cos ^{-1} x = \cos ^{-1} \left( \frac{-33}{65} \right)$.
Therefore,$x = \frac{-33}{65}$.

(B) Let $x = \cos \theta$. Since $\frac{1}{2} \leq x \leq 1$,we have $0 \leq \theta \leq \frac{\pi}{3}$.
Then,the expression becomes $\cos ^{-1}(\cos \theta) + \cos ^{-1}\left(\frac{1}{2} \cos \theta + \frac{\sqrt{3}}{2} \sin \theta\right)$.
Using the identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$,we set $\cos A = \frac{1}{2}$ and $\sin A = \frac{\sqrt{3}}{2}$,so $A = \frac{\pi}{3}$.
Thus,the expression is $\theta + \cos ^{-1}(\cos(\theta - \frac{\pi}{3}))$.
Since $0 \leq \theta \leq \frac{\pi}{3}$,then $-\frac{\pi}{3} \leq \theta - \frac{\pi}{3} \leq 0$,which implies $0 \leq \frac{\pi}{3} - \theta \leq \frac{\pi}{3}$.
Since $\cos(\theta - \frac{\pi}{3}) = \cos(\frac{\pi}{3} - \theta)$,we have $\cos ^{-1}(\cos(\theta - \frac{\pi}{3})) = \frac{\pi}{3} - \theta$.
Therefore,the expression is $\theta + (\frac{\pi}{3} - \theta) = \frac{\pi}{3}$.

(B) Given that $\tan ^{-1} x+\tan ^{-1} y+\tan ^{-1} z=\frac{\pi}{2}$.
We know the formula $\tan ^{-1} x+\tan ^{-1} y = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$.
So,$\tan ^{-1} \left( \frac{x+y}{1-xy} \right) + \tan ^{-1} z = \frac{\pi}{2}$.
Taking $\tan$ on both sides,we use $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
Let $A = \tan ^{-1} \left( \frac{x+y}{1-xy} \right)$ and $B = \tan ^{-1} z$.
Then $\tan(A+B) = \tan \left( \frac{\pi}{2} \right) = \infty$.
For $\tan(A+B)$ to be $\infty$,the denominator must be $0$.
Thus,$1 - \tan A \tan B = 0$.
$1 - \left( \frac{x+y}{1-xy} \right) z = 0$.
$1 - \frac{xz + yz}{1-xy} = 0$.
$1 - xy - xz - yz = 0$.
Therefore,$1 - xy - yz - zx = 0$.

(B) Given that,$\sin ^{-1}\left(\frac{3}{x}\right)+\sin ^{-1}\left(\frac{4}{x}\right)=\frac{\pi}{2}$.
We know that $\sin ^{-1}(A) + \cos ^{-1}(A) = \frac{\pi}{2}$,so $\sin ^{-1}\left(\frac{4}{x}\right) = \frac{\pi}{2} - \cos ^{-1}\left(\frac{4}{x}\right)$.
Rearranging the given equation: $\sin ^{-1}\left(\frac{3}{x}\right) = \frac{\pi}{2} - \sin ^{-1}\left(\frac{4}{x}\right)$.
Using the identity $\cos ^{-1}(A) = \frac{\pi}{2} - \sin ^{-1}(A)$,we get $\sin ^{-1}\left(\frac{3}{x}\right) = \cos ^{-1}\left(\frac{4}{x}\right)$.
Let $\sin ^{-1}\left(\frac{3}{x}\right) = \theta$,then $\sin \theta = \frac{3}{x}$,which implies $\cos \theta = \sqrt{1 - \left(\frac{3}{x}\right)^2} = \frac{\sqrt{x^2-9}}{x}$.
Thus,$\theta = \cos ^{-1}\left(\frac{\sqrt{x^2-9}}{x}\right)$.
Equating the two expressions: $\frac{\sqrt{x^2-9}}{x} = \frac{4}{x}$.
Squaring both sides: $x^2 - 9 = 16$,which gives $x^2 = 25$.
Since the domain of $\sin ^{-1}$ requires $|\frac{3}{x}| \le 1$ and $|\frac{4}{x}| \le 1$,we have $|x| \ge 4$. Thus,$x = 5$.

(B) We know that $\sin ^{-1} x + \cos ^{-1} x = \frac{\pi}{2}$.
Given equation: $\sin ^{-1} x - \cos ^{-1} x = \frac{\pi}{6}$.
Adding the two equations:
$(\sin ^{-1} x + \cos ^{-1} x) + (\sin ^{-1} x - \cos ^{-1} x) = \frac{\pi}{2} + \frac{\pi}{6}$
$2 \sin ^{-1} x = \frac{3\pi + \pi}{6} = \frac{4\pi}{6} = \frac{2\pi}{3}$
$\sin ^{-1} x = \frac{\pi}{3}$
$x = \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.

(C) Given $f(n) = \tan \left[\sum_{k=1}^{n} \tan ^{-1} \frac{1}{1+k(k+1)}\right]$.
We know that $\tan ^{-1} x - \tan ^{-1} y = \tan ^{-1} \left(\frac{x-y}{1+xy}\right)$.
We can write the general term as $\tan ^{-1} \left(\frac{(k+1)-k}{1+k(k+1)}\right) = \tan ^{-1}(k+1) - \tan ^{-1}(k)$.
Thus,the sum becomes a telescoping series:
$S_n = (\tan ^{-1}(2) - \tan ^{-1}(1)) + (\tan ^{-1}(3) - \tan ^{-1}(2)) + \ldots + (\tan ^{-1}(n+1) - \tan ^{-1}(n))$.
All intermediate terms cancel out,leaving $S_n = \tan ^{-1}(n+1) - \tan ^{-1}(1)$.
Using the formula $\tan ^{-1} A - \tan ^{-1} B = \tan ^{-1} \left(\frac{A-B}{1+AB}\right)$:
$S_n = \tan ^{-1} \left(\frac{(n+1)-1}{1+(n+1)(1)}\right) = \tan ^{-1} \left(\frac{n}{n+2}\right)$.
Therefore,$f(n) = \tan \left[\tan ^{-1} \left(\frac{n}{n+2}\right)\right] = \frac{n}{n+2}$.
For $n = 2021$,$f(2021) = \frac{2021}{2021+2} = \frac{2021}{2023}$.